3 variable quadratic

[chrsr34]

Hey guys.
Im having a problem with deciphering positive definite and negative definite for this quadratic form i determined from a matrix.
I dont quite understand how to evaluate if its pos def or neg def.
From what I see in my notes, q(x,y) > 0 for all x not equal to 0. This doesnt help much as with some terms being squared and some not, im not sure if q will always be positive. I also dont know if there is anything different being a 3-variable quadratic. Anyways here it is:
1. The problem statement, all variables and given/known data

q(x, y, z) = 2x² - 2xy + y² - 8yz – 2xz + 5z²

2. Relevant equations

q(x,y) > 0 for all x not equal to 0

Any help is appreciated
Thanks

Chris

[NateTG]

Have you tried applying the quadratic formula?

[D H]

What does positive definite mean in terms of the eigenvalues of the matrix?

[Dick]

Before you do a lot of work, you could also just poke around. If v=(x,y,z)=(1,1,0) then q(v,v)=1. If w=(0,1,1), q(w,w)=(-2).

[chrsr34]

hmm, well what im not sure of most is the statement "q(x,y) > 0 for all x not equal to 0"
This was given for 2 variables, i have nothing for 3 variables. So in im not even sure if we are only taking into consideration what x is or if there are 2 variables to take consideration of in 3D. This is the only definition of positive definite i have :frown:
I could do what Dick says by plug and play but im still not sure if the same definition holds for 3D....

[Dick]

For two variables the statement is q(x,y)>0 for (x,y) not equal to (0,0). For three variables the statement is q(x,y,z)>0 for (x,y,z) not equal to (0,0,0). I wrote it the way I did because I usually think of a quadratic form as a function of two vectors, like the dot product.

[D H]

To make Dick's suggestion blatantly obvious, setting x=1,y=1,z=0 yields q(x,y,z)=1 while setting x=0,y=1,z=1 yields q(x,y,z)=-2. What does that tell you regarding the question of the nature of the quadratic form?


You can rewrite the expression q(x, y, z) = 2x² - 2xy + y² - 8yz – 2xz + 5z² as the matrix expression

q(\mathbf x)
= \mathbf x\cdot(\mathbf Q \mathbf x)
= \mathbf x^T\mathbf Q \mathbf x

where \mathbf Q is a symmetric matrix and \mathbf x is the column vector

\mathbf x = \bmatrix x\\y\\z\endbmatrix

The first form (\mathbf x\cdot(\mathbf Q \mathbf x)) is apparently how Dick likes to view these forms. I prefer the second form (\mathbf x^T\mathbf Q \mathbf x). It is just a matter of preference; the two expressions are equivalent.

[Dick]

I prefer the second form as well. I think of the dot product as \mathbf x^T\mathbf I \mathbf x.

[chrsr34]

Thank you guys. It appears to me that this quad is not pos or neg definite then. This is my 3rd option.
Thanks guys

Chris